Question

while performing a myelogram, a technologist is exposed to a dose rate of 2 milligray/hour (mgy/hour) at a distance of 7 feet from the x - ray source. what is the dose rate for the radiologist standing at a distance of 0.5 feet from the source?
28 mgy/hour
0.01 mgy/hour
392 mgy/hour
0.14 mgy/hour

Explanation:

Step1: Recall the inverse - square law formula

The inverse - square law for radiation dose rate is $D_1d_1^{2}=D_2d_2^{2}$, where $D_1$ and $D_2$ are the dose rates and $d_1$ and $d_2$ are the distances from the source.

Step2: Identify the given values

We are given that $D_1 = 2$ mGy/hour, $d_1=7$ feet, and $d_2 = 0.5$ feet.

Step3: Rearrange the formula to solve for $D_2$

From $D_1d_1^{2}=D_2d_2^{2}$, we can get $D_2=\frac{D_1d_1^{2}}{d_2^{2}}$.

Step4: Substitute the values into the formula

$D_2=\frac{2\times7^{2}}{0.5^{2}}=\frac{2\times49}{0.25}=\frac{98}{0.25}=392$ mGy/hour.

Answer:

392 mGy/hour